If $p$ < 0.05,
If $p$ > 0.05
Using this parameterization where $θ_1 = µ + δ$ and $θ_2 = µ − δ$, we see that δ represents half the population difference in means, as $(θ_1 − θ_2)/2 = δ,$ and µ represents the pooled average, as $(θ_1 + θ_2)/2 = µ$. Convenient conjugate prior distributions for the unknown parameters are
$$ p(\mu, \delta, \sigma^2) = p(\mu) \times p(\delta) \times p(\sigma^2)\\ \mu \sim normal(\mu_0, \gamma_0^2)\\ \delta\sim normal(\delta_0, \tau^2_0)\\ \sigma^2\sim inverse-gamma(v_0/2, v_0\sigma^2_0/2) $$
The full conditional distribution of parameters are follows as:
$$ \left\{ \mu |y_1, y_2, \delta, \sigma^2 \right\} \sim normal(\mu_n, \gamma_n^2), \;\;where\\ \mu_n = \gamma_n^2 \times[\mu_0/\gamma_0^2+\sum_{i=1}^{n_1}(y_{i,1}-\delta)/\sigma^2 + \sum_{i=1}^{n_2} (y_{i,2} + \delta)/\sigma^2]\\\gamma_n^2=[1/\gamma_0^2+(n_1+n_2)/\gamma^2]^{-1} $$
$$ \left\{ \delta |y_1, y_2, \mu, \sigma^2 \right\} \sim normal(\delta_n, \tau_n^2), \;\;where\\ \delta_n = \tau_n^2 \times[\delta_0/\tau_0^2+\sum_{i=1}^{n_1}(y_{i,1}-\mu)/\sigma^2 + \sum_{i=1}^{n_2} (y_{i,2} + \mu)/\sigma^2]\\\tau_n^2=[1/\tau_0^2+(n_1+n_2)/\gamma^2]^{-1} $$
$$ \left\{ \sigma^2 |y_1, y_2, \mu, \delta \right\} \sim inverse-gamma(v_0/2, v_0\sigma^2_n/2), \;\;where\\ v_n = v_0 + n_1+n_2 \\v_n\sigma^2_n=v_0\sigma^2_0+\sum(y_{i,1}-[\mu+\delta])^2+\sum(y_{i,2}-[\mu,\delta])^2 $$
i) In case of $p(\mu|\cdots)$
$p(\mu|\cdots ) \propto [\prod_{i=1}^{n_1}p(y_{1,i}|\mu, \delta, \sigma^2)][\prod_{i=1}^{n_2}p(y_{2,i}|\mu, \delta, \sigma^2)p(\mu)$
$\propto [\prod_{i=1}^n\frac{1}{\sqrt 2\pi \sigma^2} exp(-\frac{1}{2\sigma^2}(y_{1,i}-\mu-\delta)^2)][\prod_{i=1}^{n_2}\frac{1}{\sqrt 2\pi \sigma^2} exp(-\frac{1}{2\sigma^2}(y_{1,i}-\mu-\delta)^2)] exp(-\frac{1}{2\gamma^2_0}(\mu-\mu_0)^2)$